Alternating Current - NEET Physics Questions
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Alternating Current

Question 1: easy

Assertion (A): If an iron rod is inserted into a steady current carrying solenoid, the current in solenoid decreases.


Reason (R): Magnetic flux linked with solenoid increases.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Inserting an iron rod increases magnetic flux \( \Phi \). By Lenz's law, this induces an opposing \( \text{EMF} \), causing current \( \text{I} \) to decrease during insertion.

Question 2: easy

Assertion (A): ac current flows through a bulb and a solenoid connected in series. If a soft iron core is inserted in the solenoid, the bulb glows much brighter.


Reason (R): The inductance of solenoid decreases on inserting soft iron core in it.


 

1. Both A & R are true and the (R) is the correct explanation of the (A)
2. Both A & R are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Inserting a soft iron core increases solenoid inductance \(L\), thus increasing inductive reactance \(X_L = omega L\). This increases circuit impedance \(Z\), reducing current \(I = V/Z\) and making the bulb dimmer. Both Assertion (A) and Reason (R) are false.

Question 3: easy

Assertion (A): A choke coil has the characteristic of high inductance and low resistance.


Reason (R): More is the inductive property of the choke coil, Power factor of the circuit approaches maximum.


 

1. Both A & R are true and the (R) is the correct explanation of the (A)
2. Both A & R are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

A choke coil has high inductance and low resistance (A is true). Power factor is \(cos\phi = R/Z = R/sqrt{R^2 + X_L^2}\). Higher inductive property (large \(X_L\)) makes \(cos\phi\) approach minimum (0), not maximum. So R is false.

Question 4: easy

Assertion (A): In a series \(LCR\) circuit at resonance, the voltage across the capacitor or inductor may be more than the applied voltage.


Reason (R): At resonance in a series \(LCR\) circuit, the voltages across inductor and capacitor are out of phase.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

At resonance, \(V_L = V_C\) but they are \(180^\circ\) out of phase. The applied voltage is \(V = IR\). Due to voltage magnification (high \(Q\) factor), \(V_L\) or \(V_C\) can be much greater than \(V\). Reason is true, but it doesn't explain *why* they can be larger than applied voltage, it explains why they cancel out to make \(V=IR\).

Question 5: easy

Assertion (A): Average power consumed in an \(AC\) circuit is equal to average power consumed by resistors in the circuit.


Reason (R): Average power consumed by capacitor and inductor is zero.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Average power in \(AC\) is \(P_{avg} = V_{rms} I_{rms} \cos\phi\). For pure inductor or capacitor, \(\phi = \pm \pi/2\) so \(cos\phi = 0\). Only resistors dissipate average power, \(P_{avg} = I_{rms}^2 R\). Hence, R correctly explains A.

Question 6: easy

Assertion (A): Peak voltage across the resistance can be greater than the peak voltage of the source in a series \(LCR\) circuit.


Reason (R): Peak voltage across the inductor can be greater than the peak voltage of the source in an series \(LCR\) circuit.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Peak voltage across resistor is \(V_R = I_0 R\). Peak source voltage is \(V_0 = I_0 Z\). Since \(Z \ge R\), \(V_R \le V_0\). So A is false. Peak voltage across inductor is \(V_L = I_0 X_L\). At resonance, if \(X_L > R\), then \(V_L\) can be greater than \(V_0\) (voltage magnification). So R is true. Thus, A is false and R is true.

Question 7: easy

Assertion (A): The power rating of an element in \(AC\) circuit refers to average power rating.


Reason (R): A given value for \(AC\) voltage or current is usually its average value.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Power rating of \(AC\) devices always refers to average power. However, \(AC\) voltage or current values (e.g., 220V) are typically Root Mean Square (RMS) values, not average values. For a full cycle, the average value of sinusoidal \(AC\) is zero.

Question 8: easy

Assertion (A): Average power consumed in a circuit is never negative.


Reason (R): Instantaneous power is always positive.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Average power consumed by passive circuits is non-negative. Instantaneous power \(p = vi\) can be negative during parts of an \(AC\) cycle, especially in reactive circuits, when energy is temporarily returned to the source.

Question 9: easy

Assertion (A): At an airport, a person is made to walk through the doorway of a metal detector.


Reason (R): Metal detector works on the principle of resonance in \(AC\) circuits.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Metal detectors use \(LC\) resonant circuits. When a metal object enters the coil's magnetic field, it changes the coil's inductance, altering the circuit's resonant frequency and triggering detection.

Question 10: easy

Assertion (A): Smaller the band width, sharper the resonance and easier it is to tune an \(LCR\) circuit.


Reason (R): Resonant frequency is arithmetic mean of half power frequencies.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Smaller bandwidth implies a higher quality factor \(Q\) and sharper resonance, leading to better frequency selectivity (easier tuning). Resonant frequency \(\omega_0\) is the *geometric mean* \(sqrt{\omega_1 \omega_2}\,\) not the arithmetic mean of half-power frequencies.